Concept of Particles and Free Body Diagram Why FBD diagrams are used during the analysis? It enables us to check the body for equilibrium. By considering the FBD, we can clearly define the exact system of forces which we must use in the investigation of any constrained body. It helps to identify the forces and ensures the correct use of equation of equilibrium. Note: Reactions on two contacting bodies are equal and opposite on account of Newton's III Law. The type of reactions produced depends on the nature of contact between the bodies as well as that of the surfaces. Sometimes it is necessary to consider internal free bodies such that the contacting surfaces lie within the given body. Such a free body needs to be analyzed when the body is deformable. Physical Meaning of Equilibrium and its essence in Structural Application The state of rest (in appropriate inertial frame) of a system particles and/or rigid bodies is called equilibrium. A particle is said to be in equilibrium if it is in rest. A rigid body is said to be in equilibrium if the constituent particles contained on it are in equilibrium. The rigid body in equilibrium means the body is stable. Equilibrium means net force and net moment acting on the body is zero. Essence in Structural Engineering To find the unknown parameters such as reaction forces and moments induced by the body. In Structural Engineering, the major problem is to identify the external reactions, internal forces and stresses on the body which are produced during the loading. For the identification of such parameters, we should assume a body in equilibrium. This assumption provides the necessary equations to determine the unknown parameters. For the equilibrium body, the number of unknown parameters must be equal to number of available parameters provided by static equilibrium condition.

1. Equivalent Force Systems
EQUIVALENT SYSTEMS for SINGLE FORCEEQUIVALENT SYSTEMS for SINGLE FORCE
Determining the effect of moving a force.
1. MOVING1. MOVING A FORCEA FORCE ONON ITS LINE OF ACTIONITS LINE OF ACTION
2. MOVING2. MOVING A FORCEA FORCE OFFOFF OF ITS LINE OF ACTIONOF ITS LINE OF ACTION

2. 1. MOVING A FORCE1. MOVING A FORCE ONON ITS LINE OF ACTIONITS LINE OF ACTION
Moving a force from A to O, when both points are on the vectors’
line of action, does not change the external effect. Hence, a force
vector is called a sliding vector. (But the internal effect of the force
on the body does depend on where the force is applied).
the two systems are equivalentthe two systems are equivalent
Equivalent Force Systems

3. 2. MOVING A FORCE2. MOVING A FORCE OFFOFF OF ITS LINE OF ACTIONOF ITS LINE OF ACTION
Moving a force from point A to B (as shown above) requires
creating an additional couple moment. Since this new couple
moment is a “free” vectorfree” vector, it can be applied at any point P on the
body.
the two systems are equivalentthe two systems are equivalent
Use this process repeatedly for systems of forcesUse this process repeatedly for systems of forces
Equivalent Force Systems

4. Moving a force from point A to B requires creating an additional
couple moment.
≡
MB= (80+40)(-150 kN)
= -18000 kN.mm
= -18.000 kN.m
Equivalent Force Systems
2. MOVING A FORCE2. MOVING A FORCE OFFOFF OF ITS LINE OF ACTIONOF ITS LINE OF ACTION
Example:

5. 2. MOVING A FORCE2. MOVING A FORCE OFFOFF OF ITS LINE OF ACTIONOF ITS LINE OF ACTION
Mo = (200 sin60°)(400 N)
= 69282 N.mm
= 69.282 N.m
Equivalent Force Systems
Example:
Determine the equivalent system
for the 400 N force acting at point O?

6. Determine equivalent force and couple at A?
Example:
Equivalent Force Systems

7. Equivalent Force Systems ‘2D’
Replace by
Equivalent
System

8. RESULTANT OF PARALLEL FORCES
AT A POINT ‘2D’
AA system of forcessystem of forces andand momentsmoments cancan bebe
simplify intosimplify into a single resultanta single resultant forceforce andand
momentmoment acting at a specified pointacting at a specified point..
Parallel system of forcesParallel system of forces

9. If the force system lies in the x-y plane (2-D case), thenIf the force system lies in the x-y plane (2-D case), then
the reduced equivalent system can be obtained usingthe reduced equivalent system can be obtained using
the following two scalar equations.the following two scalar equations.
Parallel system of forcesParallel system of forces
ARA
yR
MM
FF
∑
∑
=
=
RESULTANT OF PARALLEL FORCES
AT A POINT ‘2D’

10. Parallel system of forcesParallel system of forces
ARA
yR
MM
FF
∑
∑
=
=
RESULTANT OF PARALLEL FORCES
AT A POINT ‘2D’

11. a a a a
O
F1 F2
F3 F4 F5
y
x
Find the equivalent Force ­­­­ couple system at
O?
F1 = 100 N, F2 = 90 N,
F3 = 80 N, F4 = 70 N,
F5 = 60 N and a = 0.2 m
Parallel system of forcesParallel system of forcesExample:
RESULTANT OF PARALLEL FORCES
AT A POINT ‘2D’

12. 4 4 4
333
F1 F2 F3
h1 h2 h2h3
h3
•
P
Given that:
F1 = 20 kN, F2 = 30 kN, F3 = 40 kN and
h1 = 2 m h2 = 3 m h3 = 1 m
Replace the given system of parallel forces by a force-couple system
acting at the point P?
Parallel system of forcesParallel system of forces
Example:
RESULTANT OF PARALLEL FORCES
AT A POINT ‘2D’

13. Further Reduction ofFurther Reduction of
ParallelParallel ForceForcess ‘2D’‘2D’
Several parallel forces acting on the stick can be
replaced by a single resultant force FR acting at a
distance d from the point of grip.
The equivalent Force:
FR = F1 + F2 + .... + FN
To find distance d use:
FRd = F1d1 + F2d2 + ..... + FNdN
dN
FN
Parallel system of forcesParallel system of forces

14. ≡≡ ≡≡
Parallel system of forcesParallel system of forces
Further Reduction ofFurther Reduction of
ParallelParallel ForceForcess ‘2D’‘2D’

15. a a a a
O
F1 F2
F3 F4 F5
y
x
For the given force system find a
representative single force and its
location on the x-axis?
F1 = 100 N; F2 = 90 N; F3 = 80 N;
F4 = 70 N; F5 = 60 N; a = 0.2 m
Parallel system of forcesParallel system of forces
Example:
Further Reduction ofFurther Reduction of
ParallelParallel ForceForcess ‘2D’‘2D’

16. Example:
Replace the force system by:
1- a single force – couple resultant at point P,
2- a single force resultant along the x-axis.
Parallel system of forcesParallel system of forces
Further Reduction ofFurther Reduction of
ParallelParallel ForceForcess ‘2D’

17. Equivalent Force Systems ‘2D’
Replace by
Equivalent
System

18. AA system of forcessystem of forces andand momentsmoments cancan bebe
simplifsimplifiedied into a single resultantinto a single resultant forceforce andand
moment acting at a specified pointmoment acting at a specified point..
General system of forces and
moments
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

19. WhenWhen several forces and couple momentsseveral forces and couple moments act on aact on a
body, each forcebody, each force should beshould be movemove withwith its associatedits associated
couple moment tocouple moment to thatthat common point O.common point O.
AAdd all the forces and couple moments togetherdd all the forces and couple moments together
and findand find one resultant force-couple moment pone resultant force-couple moment pairair..
General system of forces and
moments
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

20. When several forces acting on a given system with
couple moments, a single resultant force FR can be
replaced to such a distance d along the specified
line so that, it will have the same external effect on
the given system.
i.e. Locate the resultant force FR at such a distance
d so that, this resultant force FR will overcome the
couple-moment effect also.
Determine: FR = F1 + F2 + .... + FN
Find distance d:
F d = (F d + F d + .... + F d ) + (M +M +...+M )
General system of forces and
moments
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

21. M2
M1
≡≡≡≡
General system of forces and
moments
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

22. [1] 20 points
Replace the given system of
forces and couple by a single
force and its location on the
line OE?
•
E
2m 2m
Example:
General system of forces and
moments
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

23. Example:
The beam AE is subjected to a system of coplanar forces.
Determine the magnitude, direction and the location on the
beam of a resultant force which is equivalent to the given
system of forces measured from E. i.e. Determine the
equivalent force system measured from point E.
General system of forces and
moments
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

24. C
Example: General system of forces and
moments
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

25. General system of forces and
moments
If the force system lies in the x-y plane (2-D case),
then the reduced equivalent system can be obtained
using the following three scalar equations.
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

26. M2
M1
General system of forces and
moments
≡≡
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

27. Reducing the given system of forces and couple moments
into equivalent resultant force and couple moment:
1- add all the forces algebraically,
2- determine the moments of each of these forces with respect
to that point
3- add all the existing moments the system.
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

28. For resultant moment calculations if the system contains
forces and moment then,
MMRoRo = ∑M + ∑(r X F)= ∑M + ∑(r X F)
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

29. Determine equivalent force and couple (moment) at O.
Example:
General system of forces and
moments
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

30. [1] 20 points
•E
2m 2m
Example:
General system of forces and
moments
Replace the given system of forces
and couple by equivalent force-
couple system acting at the point E.
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

31. The result obtained from r X F doesn’t depend on where
the vector r intersects the line of action of F:
r = r’ + u
r × F = (r’ + u) × F = r’ × F
because the cross product of the parallel vectors u and F
is zero.
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

32. Example: Determine the magnitude and directional sense of the
resultant moment of the forces about point O and point P.
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

33. Determine equivalent single force and couple at O.
Example:
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

34. Example:
Replace the forces acting on
the brace by an equivalent
resultant force and couple
moment acting at point A.
i.e. Determine equivalent
single force and couple
(moment) at A.
57.2°
Reducing the given system of forces and couple
moments into an equivalent SINGLE force and
couple moment at ANT POINTat ANT POINT ‘2D’

35. Example:Example:
For the given force – couple system, obtain an equivalent (SINGLE)
force along the line OP?
Reducing the given system of forces and
couple moments into an equivalent
SINGLE force ALONG A LINE ‘2D’

36. Example:
Reducing the given system of forces and
couple moments into an equivalent
SINGLE force ALONG A LINE ‘2D’
Determine equivalent single force along x-axis.

37. Example:
Replace the given forces by:
a) an equivalent resultant force and couple moment acting at point H.
b) a single force and its location acting along the line A-D-B.
10 mm
•
10 mm
• •
•
•
•
A
B
C
D
E
H
F1
F2
F3
M1
M2
F2 = 42.806 N
F1 = 50.215 N
F3 = 97.317 N
M1 = 2171.575 Nmm
M2 = 820.679 Nmm
General system of forces and
moments
Reducing the given system of forces and
couple moments into an equivalent
SINGLE force ALONG A LINE ‘2D’